Electrical Engineering: Ch 5: Operational Amp (15 of 28) The Difference Amplifier

[@essayprometheus]

Brilliant. My professor threw us that last equation for difference amplifier, but you actually explained how it works. Thank you a ton, my friend.

[@MichelvanBiezen]

You're very welcome!

[@moazelsawaf2000]

Thanks sir

[@gamjam7065]

Great video! The algebra towards the end really had me stuck for a while.

[@ezio_g]

Beautiful math work out........I tried on my own and gave up. Mine was not bad but didn't get the beautiful part at the end where we can clearly see why making the two ratios R1/Rf = R3/R4 gave us Vo = Rf/R1( V2-V1 ) which is a very easy to see how this circuit is "The Difference Amplifier". I have no problem understanding the electronic but I have to go over the math a couple of times. Nice lecture. Loved it.

[@jordanchacko3085]

best explanation ive ever seen for this topic. analog systems UD class is much easier with playlists

[@MichelvanBiezen]

Glad it was helpful!

[@corysmith4569]

First and foremost, that's a killer bow-tie. Bow-ties are cool. Secondly, you just saved me a massive headache. I was struggling through this derivation all morning. Your explanation was perfect. I do have a question about this whole circuit though.

[@ataurrahman3080]

Electric field,Electronics etc...All of your Tutorial is best. Good explanation with initial to Application. what a tutorial series.Thank u sir.#Bangladeshi EEE student.

[@zamp7927]

Rahman sab jeety raho

[@Sam-rl6ok]

Great Dr. Biezen. It is really simplified explaination. Much more simpler than my book's explaination. Best regards to you Dr.

[@holmeslaw1019]

Without deriving the math I had a hard time understanding opamps. Thank you very much.

[@أحمدالنقيب-ت3و]

Thank you very much, today you have been being my hero.

[@punsarawithanage3840]

Superb❤️ from Sri lanka🇱🇰

[@MichelvanBiezen]

Glad it is helpful. Welcome to the channel.

[@bennetbrooks1]

Thank you. Your explanation makes the topic easy to understand. What textbook do you recommend for electronics devices and op-amps.

[@themanthis837]

How do you calculate this if you add a resistor from Va to gnd ( when you want to amplify a bridge case )

[@avdallinone3798]

nice explanation sir I searching for this frm a week and now I got it from your video thank u sir

[@morto00x]

Had this exact question at a job interview. Got stuck at that very last step lol.

[@abutaymiyyahlectures]

damn i didnt know jobs ask this kind of questions :/. what job btw?

[@Mikebigmike94]

That's brutal lol, this is more algebra tricks than engineering so i hope it wasn't an engineering job. It's one thing finding the output in terms of the input but to get it into the final equation is just pure maths i think.

[@lhaellor]

What did you do at 6:45 to simplify that fraction what is the story?

[@lhaellor]

Can anyone explain or name what is this algebraic trick at this part?

[@MichelvanBiezen]

Dividing by a fraction is the same as multiplying by its inverse. 12 / (1/2) = 12 x (2/1) = 24 or (2/1) x (12) = 12 / (1/2)

[@ander1651]

Excellent! Thank you so much for your explanation!

[@adrianhove6211]

best explanation ever, thank you Sir

[@sandlertossone1813]

Can you get two different answers if you use the first circled equation and the second shorter one? My Rf/Ri=R3/R4=2 and my V1=-5 and V2=6. Using the longer equation I get 16V and the shorter I get 22.

[@lhaellor]

What did you do here at @8:13 ?

[@MichelvanBiezen]

We multiplied the quantity inside the parentheses by the fraction Rf/R1 and re-arranged the terms.

[@audar4832]

I have an question please we use this law only when resistance are equal? (R1=R4 and Rf=R3)

[@MichelvanBiezen]

They don't have to be equal, but if you do make them equal you will have that simple relationship for Vo

[@audar4832]

@@MichelvanBiezen i used the law directly to get an unkown in Nelson james reference but it is wrong the reference used th method of node voltage and get a nother solution😅🤔

[@MichelvanBiezen]

Is the setup exactly the same? Or is there a difference in the way the op amp is connected?

[@audar4832]

@@MichelvanBiezen yeh it's exactly the same

[@leonitasn9527]

so in op amp vb=vs is always the case ? I mean the difference between vb-va must ALWAYS be approx 0 ?

[@leonitasn9527]

O i got it .. it just depends if it’s connected to the ground or not .. if yes the potential is 0, but always vb=va

[@johnnycanazon]

Would it have been a bad idea to have simply jumped into using voltage devision to solve for Vb considering no current into V+ or could this cause problems in other situations?

[@MichelvanBiezen]

That should work.

[@kochuranilalu438]

Thank you sir. You made it clear👍

[@MichelvanBiezen]

Glad it helped

[@luketaylor5917]

How could you amplify the mV output from a thermocouple to a 1-5 volts signal using an op amp?

[@dachew57]

Use non-inverting Op-Amp (not Difference Amplifier). Av = 1+(Rf/R1). For example, to convert the output voltage of an analog temperature sensor, let's say a TMP35, to a 0-5v for an Arduino, set Av = 4 (Rf = 33k, R1 = 11k). Then, TMP35 maps very nicely to Arduino supply range. Of course, adjust gain for 3.3v system. TMP35: 25C = 0.250v and 125C = 1.25v. Op-Amp transfer function is f(x) = 4x, where x is TMP35 output voltage.

[@irmuunbilig8218]

thank you so much, god bless you 😇👏🙏

[@MichelvanBiezen]

Thank you. Glad you found it helpful! 🙂

[@soningadong8635]

How is the direction of current across the feedback resistance determined?

[@MichelvanBiezen]

Current flow from high potential to low potential (conventional current flow). If we don't know, we just pick a direction and if we picked the wrong direction the current will come out negative.

[@thomsonkthomas]

osm sir... understood the topic clearly...thank you..

[@luigimirabelli5504]

Great explanation! Thanks Prof!

[@2TAK322]

nice one

[@MichelvanBiezen]

Thanks

[@varnam6646]

why do we use R4?

[@geertvancompernolle2586]

I find the calculation of vo quite complex (between minute 7 and 8 something). When I do the same calculation my way, the result is vo = v2 (R4 / (R3 + R4)) ((R1 + Rf) / R1) - v1 (Rf / R1) Sorry but it's difficult to type it down like that. Anyway, by going for that shape of formula, it's easier to see what happens if R1 = R3 and Rf = R4 on one side and R1 = R3 = Rf = R4 on the other side. But again, the end result is the same (many paths lead to the same goal...). Apart from that, you're one of the very few that can really explain how a difference amplifier really works. Thumbs up!

[@agstechnicalsupport]

Thank you Michel !

[@Thabang_Kgatla1]

Thank you so much

[@MichelvanBiezen]

You are welcome.

[@paulferrer2108]

Prof. van Biezen, are we able to use superposition principle to derive the formula?

[@MichelvanBiezen]

Any method is a valid method. Some methods are easier for certain applications than others.

[@ahmedelaraby7656]

This guy is the fucking best Thanks prof

[@MuhammadUsman-lx6ye]

Sir what if R3/R4 does not equal R1/RF

[@MichelvanBiezen]

There are many different configurations for different purposes. We cover some of them in other videos. This is the configuration for the difference amplifier.

[@stevenhamer5826]

AND THATS HOW ITS DONE ! BOOM !

[@Dina-he1uc]

:D :D :D Thank you

[@MichelvanBiezen]

You are welcome. 🙂

[@jjww2788]

Thank you very much